class Solution {
public:
if((pre.size()==0)||(vin.size()==0))
return NULL;
TreeNode* root=new TreeNode(pre[0]);
int pre_index=1;
for(j=0;j<vin.size();j++)
{
if(j<i)
{
vin_left.push_back(vin[j]);
pre_left.push_back(pre[pre_index]);
pre_index++;
}
else if(j>i)
{
vin_right.push_back(vin[j]);
pre_right.push_back(pre[pre_index]);
pre_index++;
}
root->right=reConstructBinaryTree(pre_right,vin_right);
return root;
}
};
public:
TreeNode* reConstructBinaryTree(vector<int> pre,vector<int> vin) {
//1.先判断遍历的二叉树是否为空,其实一般来说检查一项就行
return NULL;
TreeNode* root=new TreeNode(pre[0]);
int i,j;
//2.中序遍历中找到根节点的位置记录下来设为i
for(i=0;i<vin.size()&&vin[i]!=pre[0];i++);
//3.根据i的位置,前序遍历的左子树存于pre_left中,右子树存于pre_right, 中序遍历同样如此
vector<int> pre_left,pre_right,vin_left,vin_right;int pre_index=1;
for(j=0;j<vin.size();j++)
{
if(j<i)
{
vin_left.push_back(vin[j]);
pre_left.push_back(pre[pre_index]);
pre_index++;
}
else if(j>i)
{
vin_right.push_back(vin[j]);
pre_right.push_back(pre[pre_index]);
pre_index++;
}
}
//4.递归调用
root->left=reConstructBinaryTree(pre_left,vin_left);root->right=reConstructBinaryTree(pre_right,vin_right);
return root;
}
};